The origin lies in the acute angle between the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ if $aa' + bb' + cc' < 0$ and:

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them: $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$.

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then:

Find the direction ratios of the normal to the plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$.

The equation of the plane passing through the point $(1, -3, -2)$ and perpendicular to the planes $x + 2y + 2z = 5$ and $3x + 3y + 2z = 8$ is:

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is: